Mysql的合并2数查询在一个查询

SELECT COUNT(*) FROM a WHERE year = 2016 && month = 5 && date = 17 && user = 1 
//return 1 row 

SELECT COUNT(*) FROM a WHERE year = 2016 && month = 5 && date = 17 && user = 2 
//return 0 row

我有2个问题，我需要检查用户1和用户2，用户1必须具有1行&用户2必须零行Mysql的合并2数查询在一个查询

我的问题是：这可能与两列合并这2个查询一起

收益为1行//return 1, 0

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2016-05-17 Benjamin W

可以使用sum()用条件来计算的条件有多少次是真的

SELECT SUM(user = 1) as u1, 
     SUM(user = 2) as u2 
FROM a 
WHERE year = 2016 and month = 5 and date = 17

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2016-05-17 10:01:53

试试这个：

SELECT SUM(user = 1) AS user1, SUM(user = 2) AS user2 
FROM a 
WHERE year = 2016 AND month = 5 AND date = 17

SELECT子句的第一个字段返回user = 1次出现次数，而第二个字段返回user = 2次出现次数。

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2016-05-17 10:01:30

是的，这就是所谓的有条件聚集：

SELECT count(CASE WHEN `user` = 1 THEN 1 END) as usr1_cnt, 
     count(CASE WHEN `user` = 2 THEN 1 END) as usr2_cnt 
FROM a 
WHERE year = 2016 and month = 5 and date = 17

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2016-05-17 10:03:09 sagi

试试这个，

SELECT 
    (
    SELECT COUNT(*) 
    FROM a 
    WHERE year = 2016 && month = 5 && date = 17 && user = 1 
    ) AS usr1, 
    (
    SELECT COUNT(*) 
    FROM a 
    WHERE year = 2016 && month = 5 && date = 17 && user = 2 
    ) AS usr2

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2016-05-17 10:04:05 Hytool

为什么从表中选择两次？这可以在一个选择中完成。 – sagi

您可以试试这个：

SELECT * from (
SELECT COUNT(*) count1 FROM a WHERE year = 2016 && month = 5 && date = 17 && user = 1) aa, 
(SELECT COUNT(*) count2 FROM a WHERE year = 2016 && month = 5 && date = 17 && user = 2) bb;

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2016-05-17 10:05:17

为什么要从表格中选择两次？这可以在一个选择中完成。 – sagi

但是我想用两张桌子来做。 –

这是一个不必要的计算！ – sagi

你也可以试试这个：

SELECT SUM(user = 1) as user1, 
     SUM(user = 2) as user2 
FROM a 
WHERE year = 2016 and month = 5 and date = 17;

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2016-05-17 10:17:54

Mysql的合并2数查询在一个查询

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